Consider my simple example below:
#include <iostream>
template <typename T>
class Base
{
public:
static constexpr int y = T::x;
};
class Derived : public Base<Derived>
{
public:
static constexpr int x = 5;
};
int main()
{
std::cout << Derived::y << std::endl;
}
In g++, this compiles fine and prints 5
as expected. In Clang, however, it fails to compile with the error no member named 'x' in 'Derived'
. As far as I can tell this is correct code. Is there something wrong with what I am doing, and if not, is there a way to have this work in Clang?
This probably isn't the answer anyone would be looking for, but I solved the problem by adding a third class:
#include <iostream>
template <typename T>
class Base
{
public:
static constexpr int y = T::x;
};
class Data
{
public:
static constexpr int x = 5;
};
class Derived : public Base<Data>, public Data {};
int main()
{
std::cout << Derived::y << std::endl;
}
It works as desired, but unfortunately it doesn't really have the benefits of CRTP!
As linked in the comments, Initializing a static constexpr data member of the base class by using a static constexpr data member of the derived class suggests that clang behaviour is standard conformant up to C++14 here. Starting with Clang 3.9, your code compiles successfully with -std=c++1z
.
An easy possible workaround is to use constexpr functions instead of values:
#include <iostream>
template <typename T>
class Base
{
public:
static constexpr int y() {return T::x();}
};
class Derived : public Base<Derived>
{
public:
static constexpr int x() {return 5;}
};
int main()
{
std::cout << Derived::y() << std::endl;
}
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With